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Question of 114

Q.(i) Find the 12th12^{th} term of the geometric progression 5,25,125,…5, 25, 125, \ldots

(2)
(ii) Find the sum to nn terms of the sequence 8,88,888,…8, 88, 888, \ldots (4)
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The GP 5,25,125,…5,25,125,\ldots has first term 55 and common ratio 55, so its 12th12^{th} term is 512=2441406255^{12}=244140625. The series 8,88,888,…8,88,888,\ldots is summed by writing each term as 88 times a repunit, converting the repunits into a geometric series in powers of 1010.

(i) 12th12^{th} term of 5,25,125,…5, 25, 125, \ldots

This is a GP with first term a=5a=5 and common ratio r=255=5r=\dfrac{25}{5}=5. The nthn^{th} term of a GP is an=arn−1a_n = ar^{n-1}, so

a12=5⋅511=512a_{12} = 5 \cdot 5^{11} = 5^{12}

Computing: 52=25, 54=625, 58=3906255^2=25,\ 5^4=625,\ 5^8=390625, so 512=58⋅54=390625×625=2441406255^{12}=5^8\cdot5^4=390625\times625=244140625.

a12=244140625a_{12} = 244140625

(ii) Sum to nn terms of 8,88,888,…8, 88, 888, \ldots

Factor out 88 from every term:

Sn=8+88+888+⋯(n terms)=8(1+11+111+⋯(n terms))S_n = 8+88+888+\cdots \text{(}n\text{ terms)} = 8\big(1+11+111+\cdots \text{(}n\text{ terms)}\big)

Multiply and divide the bracket by 99 so each term becomes 10k−110^k-1: …

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